Question

Calculate the value of δs when 64.0 g of ga(l) solidifies at 29.8 ∘c.

Answer 1

The variation of entropy of a substance is given by
$\delta S = \frac{\delta Q}{T}$ (1)
where
$\delta Q$ is the heat exchanged in the process
T is the absolute temperature at which the transformation occurs.

The process in the problem is the solidification of the liquid Gallium, which releases an amount of heat equal to:
$\delta Q = m L_f$
where m is the mass of the substance and $L_f = 80.1 J/g$ is the latent heat of fusion of Gallium. Using m=64.0 g, we find
$\delta Q= m L_f = (64.0 g)(80.1 J/g)=-5126.4 J$
where the negative sign means the Gallium is releasing heat to the environment.

Now we can use equation (1) to find the variation of entropy, but first we need to convert the temperature into Kelvin:
$T=29.8^{\circ} + 273.15 = 302.95 K$

And so the variation of entropy is 
$\delta S = \frac{\delta Q}{T}= \frac{-5126.4 J}{302.95 K}=-16.92 J$
and the negative sign means the entropy in the process is decreasing.